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How to integrate the following?

$$\int\frac{\sqrt{(1-x^2)^3}}{x^6} \;dx .$$

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Let $x = \sin(\theta)$. We then get $dx = \cos(\theta) d \theta$. Hence, $$I= \int \frac{\cos^3(\theta)}{\sin^6(\theta)} \cos(\theta) d \theta = \int \cot^4(\theta) cosec^2(\theta) d \theta$$

Let $\cot(\theta) = t$, then $-cosec^2(\theta) d \theta = dt$. Hence, $$I = -\int t^4 dt = -\frac{t^5}{5} + c = -\frac{\cot^5(\theta)}{5} + c = -\frac15 \left( \frac{\sqrt{1-x^2}}{x} \right)^5 + c$$

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+1: Your answer hadn't appeared yet when I was typing my comment. Or, at the very least, my eyes didn't see it. – Joe Johnson 126 Oct 30 '11 at 16:18
There is a minor mistake out over there:If $t=cot\theta$ then $dt=-cosec^2\theta d\theta$ – Anamitra Palit Dec 4 '11 at 4:07
@AnamitraPalit: Thanks for pointing that out. – user17762 Dec 4 '11 at 5:54

Hint: Do the substitution $x = \sin(\alpha)$, $\alpha \in [-\frac{\pi}{2},\frac{\pi}{2}]$. Then you get $$\int \frac{\cos^4(\alpha)}{\sin^6(\alpha)} d\alpha.$$ Now consider the following identities: $$D\left(\frac{\cos^3(\alpha)}{\sin^5(\alpha)}\right) = \frac{-3\cos^2(\alpha)\sin^6(\alpha) - 5\sin^4(\alpha)\cos^4(\alpha)}{\sin^10(\alpha)} = -3\frac{\cos^2(\alpha)}{\sin^4(\alpha)} - 5 \frac{\cos^4(\alpha)}{\sin^6(\alpha)}$$ $$D\left(\frac{\cos(\alpha)}{\sin^3(\alpha)}\right) = \frac{-\sin^4(\alpha)-3\sin^2(\alpha)\cos^2(\alpha)}{\sin^6(\alpha)} = -\frac{1}{\sin^2(\alpha)} - 3 \frac{\cos^2(\alpha)}{\sin^4(\alpha)}$$

By using these it boils down to being able to integrate $\frac{1}{\sin^2(\alpha)}$.

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Alternatively, you can rewrite $\frac{\cos^4(\alpha)}{\sin^6(\alpha)}$ as $\cot^4(\alpha)\csc^2(\alpha)$ and use a $u$-substitution for $u=\cot(\alpha)$, $du=-\csc^2(\alpha)\,d\alpha$. Then you need only integrate $-\int u^4\, du$. – Joe Johnson 126 Oct 30 '11 at 16:14
@Joe Johnson 126: Yes, you're right. It seems to be what Sivaram Ambikasaran did in his answer. And it's simpler. – J. J. Oct 30 '11 at 17:43


$$ \begin{align*} I &= \int \frac{\sqrt{(1-x^2)^{3}}}{x^6} dx \\ &= \int \left( \frac{1-x^2}{x^2} \right)^{3/2} \frac{1}{x^3}dx \\ &= \int \left(\frac{1}{x^2}-1 \right)^{3/2} \cdot \frac{1}{x^3} dx \end{align*} $$

Make the substitution: $z= \frac{1}{x^2}-1$, so that $dz=\frac{-2}{x^3} ~dx$.

$$ \begin{align*} \text{Integral} &= -\frac{1}{2}\int z^{3/2} dz \\ &= -\frac{1}{2} \cdot \frac{2}{5}z^{5/2}+C \\ &= -\frac{1}{5} \left(\frac{1}{x^2} - 1 \right)^{5/2}+C \\ &= - \frac{1}{5} \left(\frac{\sqrt{1-x^2}}{x} \right)^5+C. \end{align*} $$

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